User:Melissa Novy/Notebook/CHEM-571/2012/11/13

(Difference between revisions)

Revision as of 15:09, 13 November 2012

50 mL of 0.05 M sodium phosphate buffer was made with deionized water. The pH of the solution was adjusted to 7.4 using 3 drops of 12 M HCl.

5 mL of 0.1 mM adenosine was made.
- 100 mg of adenosine was dissolved in 1 mL of 0.05 M phosphate buffer, giving a solution of 0.00373 M adenosine.
- 0.134 mL of 0.00373 M adenosine was diluted to a final volume of 5 mL with 0.05 M phosphate buffer, giving a solution of 0.1 mM adenosine.

  (0.001 g)/(268.02 g/mol) = 3.73×10^-6 mol
  (3.73×10^-6 mol)/0.001 L = 0.00373 M
  (0.00373 M)(V₁)=(5 mL)(0.1×10^-3 M)
  V₁ = 0.134 mL

@@ Line 15: / Line 15: @@
 <!-- ##### DO NOT edit above this line unless you know what you are doing. ##### -->
-==Entry title==
+==Objectives==
-* Insert content here...
+* Analyze Au-HRP solutions with UV-vis spectroscopy.
+* Analyze solutions of Au-HRP and Au-lysozyme with AAS.
+* Conduct ADA activity assay on ADA fractions using adenosine.
+==ADA Activity Assay==
+* 50 mL of 0.05 M sodium phosphate buffer was made with deionized water.  The pH of the solution was adjusted to 7.4 using 3 drops of 12 M HCl.
+* 5 mL of 0.1 mM adenosine was made.
+** 100 mg of adenosine was dissolved in 1 mL of 0.05 M phosphate buffer, giving a solution of 0.00373 M adenosine.
+** 0.134 mL of 0.00373 M adenosine was diluted to a final volume of 5 mL with 0.05 M phosphate buffer, giving a solution of 0.1 mM adenosine.
+   '''(0.001 g)/(268.02 g/mol) = 3.73×10<sup>-6</sup> mol'''
+   '''(3.73×10<sup>-6</sup> mol)/0.001 L = 0.00373 M'''
+   '''(0.00373 M)(V<sub>1</sub>)=(5 mL)(0.1×10<sup>-3</sup> M)'''
+   '''V<sub>1</sub> = 0.134 mL'''